$\displaystyle \left(i\right)\\[2ex] \displaystyle L^{-1}\ \frac{se^{-\pi{}s}}{s^2+3s+2} \\[3ex] \displaystyle Let\ \ \varphi{}(s)\ =\ \frac{s}{s^2+3s+2} \\[2ex] \displaystyle L[f(t)]\ =\ \varphi{}(s) \\[2ex] \begin{align*} \displaystyle f(t)\ &=\ L^{-1}\ \varphi{}(s) \\[2ex] \displaystyle &=\ L^{-1}\ \frac{s}{s^2+3s+2} \\[2ex] \displaystyle &=\ L^{-1}\ \frac{s}{\left(s^2+3s+\frac{9}{4}\right)+\ \left(-\frac{9}{4}+2\right)} \\[2ex] \displaystyle &=L^{-1}\ \frac{s}{{\left(s+\frac{3}{2}\right)}^2-\ \left(\frac{1}{4}\right)} \\[2ex] \displaystyle &={\ L}^{-1}\ \frac{s+\frac{3}{2}-\frac{3}{2}}{{\left(s+\frac{3}{2}\right)}^2-\ {\left(\frac{1}{2}\right)}^2} \\[2ex] \displaystyle &={\ L}^{-1}\ \frac{s+\frac{3}{2}}{{\left(s+\frac{3}{2}\right)}^2-\ {\left(\frac{1}{2}\right)}^2}-\frac{3}{2}{\ L}^{-1}\ \frac{1}{{\left(s+\frac{3}{2}\right)}^2-\ {\left(\frac{1}{2}\right)}^2} \\[2ex] \displaystyle &=\ e^{\frac{-3t}{2}}{\ L}^{-1}\ \frac{s}{{\left(s\right)}^2-\ {\left(\frac{1}{2}\right)}^2}\ -\ \frac{3}{2}\ e^{\frac{-3t}{2}}{\ L}^{-1}\ \frac{1}{{\left(s\right)}^2-\ {\left(\frac{1}{2}\right)}^2} \\[2ex] \displaystyle &=e^{\frac{-3t}{2}}\cosh\ \frac{t}{2}\ –\ 2.\ \frac{3}{2}\ e^{\frac{-3t}{2}}\ \sin h\ \frac{t}{2} \\[2ex] \displaystyle &=\ e^{\frac{-3t}{2}}\left(\cos h\ \frac{t}{2}-\ 3\sin h\ \frac{t}{2}\right) \\[2ex] \displaystyle L^{-1}\ \left[e^{-as}\phi\left(s\right)\right]&=f\left(t-a\right)H\left(t-a\right) \\[2ex] \displaystyle {\therefore{}\ L}^{-1}\ \left[e^{-\pi{}s}\phi\left(s\right)\right]\ &=\ L^{-1}\ \frac{se^{-\pi{}s}}{s^2+3s+2} \\[2ex] &= e^{\frac{-3\left(t-\pi{}\right)}{2}}\left(cosh\ \frac{t-\pi{}}{2}-\ 3sinh\ \frac{t-\pi{}}{2}\right)\ H(t-\pi{})\\[2ex] \end{align*}\\ $

$ \displaystyle (ii)\\[2ex] \displaystyle \ L[t.\ H(t\ -\ 4)+t^2\ \delta{}(t-4)] \\[3ex] $

$\displaystyle =\ L[t.\ H(t\ -\ 4)]\ +\ L[t^2\ \delta{}(t-4) \\[3ex] $

$\displaystyle For\ \ \ L[t.\ H(t\ -\ 4)] \\[2ex] $

$\displaystyle L[f(t).H(t\ -\ a)]\ =\ e^{-as}\ L[f(t+a)] \\[2ex] $

$ \displaystyle Here\ \ f(t)\ =\ t \\[2ex] \displaystyle f(t+a)\ =\ t+4 \\[2ex] $

$\displaystyle L[f(t+a)]\ =\ \frac{1}{s^2}+\ \frac{4}{s} \\[2ex] \displaystyle \therefore{}\ L[t.\ H(t\ -\ 4)]\ =\ e^{-4s}\ \left(\frac{1}{s^2}+\frac{4}{s}\right) \\[2ex] \displaystyle For\ \ \ L[t^2\ \delta{}(t-4) \\[2ex] $

$\displaystyle L[f(t)\delta{}(t-a)]\ =\ e^{-as}\ f(a) $

$ \displaystyle Here\ \ f(t)\ =\ t^2 \\[2ex] \displaystyle a\ =\ 4 \\[2ex] \displaystyle f(a)\ =\ f(4)\ =16 \\[2ex] \displaystyle \therefore{}\ \ L[t^2\ \ \delta{}(t-4)]\ =\ e^{-4s}\ f(4)\ =16\ e^{-4s} \\[2ex] \displaystyle L[t.\ H(t\ -\ 4)+t^2\ \ \delta{}(t-4)] \\[2ex] \displaystyle =\ L[t.\ H(t\ -\ 4)]\ +\ L[t^2\ \ \delta{}(t-4)] \\[2ex] \displaystyle =\ e^{-4s}\left(\frac{1}{s^2}+\ \frac{4}{s}\right)+\ 16\ e^{-4s} \\[2ex] \displaystyle =e^{-4s}\left(\frac{1}{s^2}+\ \frac{4}{s}+16\right) \\[2ex] $